Numerical methods

CIE A-Level Mathematics· 15 min read

1. Locating Roots by Sign Change★★☆☆☆⏱ 15 min

If a function $f (x)$ is continuous on the closed interval $[a, b]$ , and $f (a)$ and $f (b)$ have opposite signs, then there is at least one root of $f (x) = 0$ in the open interval $(a, b)$ . This is the foundation of all numerical root finding.

📘 Definition

Root of an equation

A value $x = α$ such that $f (α) = 0$ , also called a zero of the function $f$ .

📐 Worked Example

Show that $e^{x} - 3 x^{2} = 0$ has a root between $x = 3$ and $x = 4$ .

1
Define the function for the equation:
2
$f(x) = e^x - 3x^2$
3
Evaluate $f (x)$ at the endpoints of the interval:
4
$f(3) = e^3 - 3(3)^2 \approx 20.0855 - 27 = -6.9145 < 0$
5
$f(4) = e^4 - 3(4)^2 \approx 54.598 - 48 = 6.598 > 0$
6
$f (x)$ is continuous for all real $x$ , as a combination of exponential and polynomial functions. Since $f (3)$ and $f (4)$ have opposite signs, there is at least one root in $(3, 4)$ .

Exam tip:

Always explicitly state that the function is continuous when justifying a root in an interval, CIE allocates one mark for this step.

2. Fixed Point Iteration★★★☆☆⏱ 20 min

To find an approximate value for a root, we rearrange $f (x) = 0$ into the form $x = g (x)$ . Starting from an initial guess $x_{0}$ , we generate a sequence of approximations using the recurrence relation $x_{n + 1} = g (x_{n})$ .

📘 Definition

Convergence Condition

If $∣ g^{'} (α) ∣ < 1$ (where $α$ is the fixed point), the iteration will converge to $α$ for starting values close enough to $α$ . If $∣ g^{'} (α) ∣ > 1$ , the iteration will diverge.

📐 Worked Example

Find the root of $x^{3} - 3 x + 1 = 0$ between 0 and 1, correct to 2 decimal places, starting from $x_{0} = 0.5$ .

1
Rearrange the equation into $x = g (x)$ form:
2
$x^3 + 1 = 3x \implies x = \frac{x^3 + 1}{3}, \quad g(x) = \frac{x^3 + 1}{3}$
3
Check convergence: $g^{'} (x) = x^{2}$ , so for $0 < x < 1$ , $∣ g^{'} (x) ∣ < 1$ , so iteration converges.
4
Generate the sequence of approximations:
5
$x_1 = \frac{(0.5)^3 + 1}{3} = 0.375$
6
$x_2 = \frac{(0.375)^3 + 1}{3} \approx 0.35095$
7
$x_3 \approx 0.34774, \quad x_4 \approx 0.34730$
8
Both $x_{3}$ and $x_{4}$ round to 0.35 to 2 decimal places, so the root is approximately 0.35.

Exam tip:

Always show two successive iterations that round to the same value to confirm your accuracy is correct.

3. Newton-Raphson Method★★★☆☆⏱ 20 min

The Newton-Raphson method is a fast, tangent-based iterative method that uses differentiation to improve approximations. It is one of the most commonly tested methods in CIE P3 exams.

📘 Definition

Newton-Raphson Iteration

A method to find successive approximations for a root of $f (x) = 0$ , given by the formula:

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

📐 Worked Example

Find the positive root of $x^{2} - 5 = 0$ correct to 3 decimal places, starting from $x_{0} = 2$ .

1
We have $f (x) = x^{2} - 5$ , so $f^{'} (x) = 2 x$ . Substitute into the Newton-Raphson formula:
2
$x_{n+1} = x_n - \frac{x_n^2 - 5}{2x_n} = \frac{x_n^2 + 5}{2x_n}$
3
First iteration with $x_{0} = 2$ :
4
$x_1 = \frac{2^2 + 5}{2(2)} = 2.25$
5
Second iteration:
6
$x_2 = \frac{(2.25)^2 + 5}{2(2.25)} \approx 2.23611$
7
Third iteration:
8
$x_3 \approx 2.23607$
9
Both $x_{2}$ and $x_{3}$ round to 2.236 to 3 decimal places, so the root is approximately 2.236.

Exam tip:

If you are asked to use Newton-Raphson, always write the full formula before substituting values to get full method marks.

4. Bisection Method★★☆☆☆⏱ 15 min

The bisection method is a simple iterative method based on the sign change principle. It is slower than other methods but guaranteed to converge if a root exists in the initial interval.

📐 Worked Example

Find the root of $e^{x} - 3 x^{2} = 0$ between 3 and 4, correct to 1 decimal place.

1
We know $f (3) < 0$ , $f (4) > 0$ . Find the midpoint of the interval and check the sign of $f$ at the midpoint:
2
$f(3.5) \approx -3.635 < 0 \implies \text{root in } (3.5, 4)$
3
$f(3.75) \approx 0.3325 > 0 \implies \text{root in } (3.5, 3.75)$
4
$f(3.625) \approx -1.75 < 0 \implies \text{root in } (3.625, 3.75)$
5
$f(3.6875) \approx -0.73 < 0 \implies \text{root in } (3.6875, 3.75)$
6
All values in $(3.6875, 3.75)$ round to 3.7 to 1 decimal place, so the root is approximately 3.7.

Exam tip:

Record the sign of f(x) at each midpoint to avoid mixing up which interval to keep next.

5. Common Pitfalls

Wrong move:

Forgetting to state the function is continuous when justifying a root in an interval

Why:

CIE explicitly allocates one mark for this statement, which is often lost

Correct move:

Always add the line 'f(x) is continuous on [a, b]' after calculating f(a) and f(b)

Wrong move:

Stopping after one iteration when asked for a given accuracy

Why:

Examiners require you to confirm your approximation is correct to the required accuracy

Correct move:

Calculate at least two successive approximations that round to the same value

Wrong move:

Claiming no root exists when there is no sign change in an interval

Why:

Tangent roots and pairs of roots do not produce a sign change but still exist

Correct move:

Only use sign change to confirm a root exists, never to confirm a root does not exist

Wrong move:

Using Newton-Raphson when $f^{'} (x_{n})$ is very close to zero

Why:

This causes near division by zero, leading to a divergent or incorrect iteration

Correct move:

Change your starting approximation or use the bisection method instead

6. Quick Reference Cheatsheet

Method	Key Rule	Convergence	Accuracy Check
Sign Change	f(a)f(b) < 0 + continuous = root in (a,b)	N/A (location)	Check interval bounds round to same value
Fixed Point Iteration	$x_{n + 1} = g (x_{n})$ , converges if $∣ g^{'} (α) ∣ < 1$	Linear	Two iterations round to same value
Newton-Raphson	$x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}$	Quadratic (fast)	Two iterations round to same value
Bisection	Halve interval, keep subinterval with sign change	Linear (slow, guaranteed)	Check interval bounds round to same value

7. Frequently Asked

Do I need to prove convergence in CIE exams?

Most questions only ask you to locate a root or apply an iterative method. You only need to check $∣ g^{'} (x) ∣ < 1$ if explicitly asked, but it is good practice to mention it.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 12
10 mark root finding question
2023 · 13
Newton-Raphson method question
2021 · 22
Iterative convergence check

Going deeper

syllabusOfficial CIE 9709 P3 Syllabus

What's Next

Numerical methods are the foundation for solving real-world problems in engineering, data science and physics, where most equations cannot be solved algebraically. The iterative principles you learned here for root finding extend directly to numerical integration, differential equations and other advanced numerical techniques. This subtopic relies heavily on differentiation, so revisiting that topic will help you master harder numerical problems that appear in CIE P3 exams.